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Some Basic Concepts in Chemistry NEET Questions — Free MCQ Practice

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Which of the following pairs of gases contains the same number of molecules:-

A.

16g of O₂ and 14g of N₂

B.

8g of O₂ and 22g of CO₂

C.

28g of N₂ and 22g of CO₂

D.

32g of O₂ and 32g of N₂

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Explanation

Mole (n) = $\frac{W e i g h t}{M o l e c u l a r w e i g h t} = \frac{N o . o f m o l e c u l e s}{N_{A}}$

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The number of ions present in 2L of a solution of 1.6M K₄[Fe(CN)₆] is:-

A.

4.8 x 10²²

B.

4.8 x 10²³

C.

9.6 x 10²⁴

D.

9.6 x 10²²

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Explanation

No. of molecules of K₄[Fe(CN)₆] = 1.6 x 2 x 6 x 10²³

No. of ions = 5 x 1.6 x 2 x 6 x 10²³

= 9.6 x 10²⁴

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The mass of CO₂ that shall be obtained by heating 10 kg of 80% pure lime stone (CaCO₃) is:-

A.

4.4 kg

B.

6.6 kg

C.

3.52 kg

D.

8.8 kg

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Explanation

CaCO₃(S) $\overset{∆}{\to}$ CO₂(g) + CaO(s)

100 kg $\to$ 44 kg

$\frac{80}{100} \times 10 k g$ $\to$ $\frac{44}{100} \times \frac{80}{100} \times 10 = 3.52 k g$

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When 1L of CO₂ is heated with graphite,the volume of the gases collected is 1.8L. What will be the number of moles of CO produced at STP?

A.

0.0357

B.

0.0714

C.

0.0803

D.

14

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Explanation

The reaction is: C(s) + CO2(g) â†’ 2CO(g). Initially, there are 1 mol of CO2. After the reaction, the total volume is 1.8 L, which corresponds to 1.8 mol of gaseous product (assuming STP). Since 1 mol of CO2 produces 2 mol of CO, the moles of CO produced is 0.0714 mol.

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At STP the moles of oxygen atoms in 2.8 L of CO₂ gas is:-

A.

0.125

B.

0.25

C.

0.5

D.

1

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Explanation

Moles of CO₂ = $\frac{v o l u m e o f g a s}{22.4} = \frac{2.8}{22.4}$ =0.125 mol

moles of oxygen atom in given CO₂ = 0.125 x 2=0.25 mol

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Which of the following relation is incorrect:-

A.

1 mol N^-3 ions have 10 mole electrons

B.

1 mol O^-2 ions have 6.023 x 10²² electrons

C.

1 mol CH₄ has 10 mole protons

D.

1 mol water has 10 mole protons

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Explanation

1 mol N^-3 ion $\to$ 1 x 10 mol e^-

1 mol O^-2 ion $\to$ 1 x 10 mol e^-= 6.023x10^-24e^-

1 mol CH₄ $\to$ 1 x 10 mol proton

1 mol H₂O $\to$ 1 x 10 mol proton

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Mass of sucrose C₁₂H₂₂O₁₁ produced by mixing 84 gm of carbon, 12gm of hydrogen and 56L. O₂ at 1atm & 273 K according to give reaction, is C(s) + H₂(g) +O₂(g) $\to$ C₁₂H₂₂O₁₁ (s)

A.

138.5

B.

155.5

C.

172.5

D.

199.4

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Explanation

Balanced Equation:

24C + 22H₂ + 11O₂ -> 2C₁₂H₂₂O₁₁

moles for reaction:

7(84/12), 6(12/2), (56/22.4)

dividing by coefficients in balanced equation & comparing:

7/24, 6/22, (56/22.4)/11

0.29, 0.27, 0.22

Oxygen is the limiting reagent. So, weight of sucrose obtained = 0.22 * 2(coefficient for sucrose in reaction) * 342(sucrose molecular weight) = 155.5g

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A mixture contains Na₂CO₃ and NaHCO₃ and wt. of mixture is 10gm. Mixture on heating liberates 56ml of CO₂ at S.T.P, wt of Na₂CO₃ in mixture is

A.

9.58 gm

B.

5 gm

C.

7.5 gm

D.

8 gm

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Explanation

Only NaHCO₃evolve CO₂ gas. Na₂CO₃ does not evolve CO₂ gas even on red heating.

So, the only reaction happening is:

2NaHCO₃ -> Na₂CO₃ + CO₂ + H₂O

Weight:

2(23+1+12+16*3) -> (2*23+12+16*3) + (12+16*2) + (1*2+16)

168g -> 102g + 44g + 18g

Also, CO₂mass as per 56ml volume obtained at STP = 56/(22.4 * 1000) * (44) = 0.11g

So, NaHCO₃weight = 0.11/44*168 = 0.42g (using ratio of weights from equation)

Therefore, original Na₂CO₃ weight in the original mixture = 10 - 0.42 = 9.58g

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How many moles of KMnO₄ are needed a mixture of 1 mole of each FeSO₄ & FeC₂O₄ in acidic medium

A.

$\frac{4}{5}$

B.

$\frac{5}{4}$

C.

$\frac{3}{4}$

D.

$\frac{5}{3}$

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Explanation

Equivalents of KMnO₄ = equivalent of FeSO₄ + equivalent of FeC₂O₄

x × 5=1 × 1 + 1 × 3

x = $\frac{4}{5}$ mole

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The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to, an excess of potassium iodide solution liberating iodine according to the following equation 2Cu² (aq) + 4I^– (aq) $\overset{}{⇌}$ 2CuI(s) + I₂(aq) The iodine liberated required 24.5cm³ of a 0.100 mole dm^-3 solution of sodium thiosulphate 2S₂O₃^2- (aq) + I₂(aq) $\to_{}^{}$ 2I^– (aq) + S₄O₆^2- (aq) the percentage of copper, by mass in the copper(ll) salt is. [Atomic mass of copper = 63.5]

A.

64.2

B.

51.0

C.

48.4

D.

25.5

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Explanation

From given reactions

mmoles of hypo = mmoles of iodine × 2

= mmoles of Cu²⁺ ions

= 24.5 × 0.1 mmoles

So mass of copper = 24.5 × 0.1 × 10^–3 × 63.5 gm

So % of copper = $\frac{24 .5 \times 0 .1 \times 10^{- 3} \times 63 .5}{0 .305} \times 100 % ≃ 51 .0 %$

Some Basic Concepts in Chemistry MCQs for NEET — Chemistry Questions with Answers

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